# Today's calculation of integral #9

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• Apr 12th 2008, 05:22 PM
Krizalid
Today's calculation of integral #9
Find $\displaystyle \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.$
• Apr 12th 2008, 08:24 PM
Mathstud28
Ok
Quote:

Originally Posted by Krizalid
Find $\displaystyle \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.$

You can rewrite it as $\displaystyle \int_0^1{\sum_{n=0}^{\infty}x^{n+2}}\cdot\bigg(-\sum_{n=0}^{\infty}\frac{x^{n}}{n}\bigg)^{-2}$...which then is equal to$\displaystyle \int_0^1{\sum_{n=0}^{\infty}x^{n+2}\cdot{2\sum_{n= 0}^{\infty}\sum_{n=0}^{\infty}\frac{n^2}{x^{2n}}}}$...and then $\displaystyle 2\int_0^1{\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\s um_{n=0}^{\infty}\frac{x^{2-n}}{n}}$...so wed get $\displaystyle 2\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\sum_{n=0}^ {\infty}\frac{x^{3-n}}{n(3-n)}$ evaluated from 0 to 1...I know I know I did it completely wrong...but will someone point out where? I just did this for novelty (Rofl)
• Apr 13th 2008, 05:20 AM
Krizalid
Quote:

Originally Posted by Mathstud28
I just did this for novelty (Rofl)

Then just don't do it if you don't know how to tackle what it comes.

It does exist a quick solution.
• Apr 13th 2008, 08:01 AM
ThePerfectHacker
I did not solve it yet. Here is the first step I was thinking. Let $\displaystyle t = -\ln (1-x)$.
This gives us, $\displaystyle \int_0^{\infty} \left( \frac{1 - e^{-t}}{t} \right)^2 dt$.
• Apr 13th 2008, 08:04 AM
Moo
Hello,

Let $\displaystyle u=\ln(1-x)$

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$\displaystyle du=-\frac{dx}{1-x}$

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$\displaystyle e^u=1-x$

$\displaystyle x=1-e^u$

$\displaystyle x^2=(1-e^u)^2$

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If x=0 then $\displaystyle u=0$

If x=1 then $\displaystyle u=-\infty$

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Inverting the extremities of the integral with the - sign coming from du

$\displaystyle \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)} dx = \int_{-\infty}^0 \frac{(1-e^u)^2}{u^2} du$

Where did i go wrong ? :'(
• Apr 13th 2008, 08:08 AM
ThePerfectHacker
Quote:

Originally Posted by Moo
Where did i go wrong ? :'(

You did not solve it, look at my post above. This only changes the integral into a (probably) simpler one. Now you need to work with this new integral.
• Apr 13th 2008, 08:11 AM
Moo
Actually, i've been trying but gave up an hour ago... The first parts of integration by parts were divergent oO

And ok, when i posted i saw the first version of your integral and got anxious because it was not the same...
• Apr 13th 2008, 08:23 AM
Mathstud28
So is it this?
Quote:

Originally Posted by ThePerfectHacker
I did not solve it yet. Here is the first step I was thinking. Let $\displaystyle t = -\ln (1-x)$.
This gives us, $\displaystyle \int_0^{\infty} \left( \frac{1 - e^{-t}}{t} \right)^2 dt$.

is that then equal to $\displaystyle \int{\bigg(-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n-1}}{n!}+\sum_{n=0}^{\infty}\frac{(-2)^{n}x^{n-2}}{n!}+1\bigg)dx}$$\displaystyle =-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n}}{n\cdot{n!}}+\sum_{n=0}^{\infty}\frac{ (-2)^{n}x^{n-1}}{(n-1)n!}+x$ evaluated from 0 to ∞?
• Apr 13th 2008, 08:33 AM
Moo
Yop,

Where did $\displaystyle \frac{1}{t^2}$ go ? :p (it's it, instead of +1)

But your method doesn't simplify the problem... or i've missed a point ^^
• Apr 13th 2008, 08:38 AM
Mathstud28
oops
Quote:

Originally Posted by Moo
Yop,

Where did $\displaystyle \frac{1}{t^2}$ go ? :p (it's it, instead of +1)

But your method doesn't simplify the problem... or i've missed a point ^^

what I said except change x to $\displaystyle \frac{-1}{x}$
• Apr 13th 2008, 10:26 AM
Krizalid
Quote:

Originally Posted by ThePerfectHacker
I did not solve it yet. Here is the first step I was thinking. Let $\displaystyle t = -\ln (1-x)$.

This is a good start. :D

Now find the other one.
• Apr 16th 2008, 06:22 PM
Krizalid
Hint: construct a double integral. (Following on post #4.)
• Apr 26th 2008, 12:06 PM
Krizalid
What we need is $\displaystyle \frac{1}{t^{2}}=\int_{0}^{\infty }{xe^{-tx}\,dx},\,t>0.$ From there you may observe that the conclusion follows.
• Apr 27th 2008, 04:45 PM
ThePerfectHacker
Quote:

Originally Posted by Krizalid
What we need is $\displaystyle \frac{1}{t^{2}}=\int_{0}^{\infty }{xe^{-tx}\,dx},\,t>0.$ From there you may observe that the conclusion follows.

Thank you for that. I had a similar idea: $\displaystyle \frac1{t^2} = \int_0^{\infty} e^{-t^2x} dx$ but that gave a nasty integral.
• Apr 29th 2008, 08:39 AM
ThePerfectHacker
In general, $\displaystyle \frac{(n-1)!}{t^n} = \int_0^{\infty} x^{n-1} e^{-tx} dx$ for $\displaystyle t>0$ and $\displaystyle n\geq 1$. This seems to be a useful fact to know in double integration.
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